are used for jewelry and coinage)—choice (E).
- B There are several ways to approach this one. Again, process of
elimination is very useful here. Chlorine (choice (A)) and nitrogen
dioxide (choice (E)) gases are colored—greenish and orange,
respectively. Therefore, they cannot be the gas in question. Furthermore,
it’s difficult to choose carbon dioxide (choice (D)) because there are no
carbon atoms anywhere in this experiment. Finally, as a rule of thumb, a
colorless gas produced from reactions between metals and acids is
hydrogen—choice (B).
- E This requires a bit of general chemistry knowledge. The colors of the
gases are
Cl 2 —green
H 2 —colorless
O 2 —colorless
CO 2 —colorless
NO 2 —orange/brown
Choice (E) is the best choice.
- B First, the neutralization reaction that occurs here is
KOH + HI → KI + H 2 O
The trick is to realize that the number of K’s and I’s are not
changing during the reaction; but since the solutions are
being added, the volume is doubling. If the volume doubles,
then the initial solution concentrations (0.2 M) are halved
(0.1 M).
- D An O2− ion has gained 2 electrons to fill its outer shell. This gives an
O2− ion the same electron configuration as Ne. That is, the 2 are
isoelectronic.
